Hard Further Maths Vertical circular motion Questions

Challenging, exam-style Further Maths Vertical circular motion questions with worked solutions. Stretch yourself on the hardest vertical-circular-motion, complete-circles, symbolic, radial-equation problems.

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Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
A particle PP of mass mm is attached to one end of a light rigid rod of length aa, freely pivoted at a fixed point OO, and is projected horizontally with speed uu from the point vertically below OO. Which of the following best describes the motion when 4ga<u2<5ga4ga<u^{2}<5ga?
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Worked solution

  1. Check that the circle is completed

    u2>4ga ⇒ vtop2=u2−4ga>0u^{2}>4ga\ \Rightarrow\ v_{\text{top}}^{2}=u^{2}-4ga>0

    A rod needs only u2>4gau^{2}>4ga, so PP does get round.

  2. Check what a string would do

    u2<5ga ⇒ Ttop=mu2a−5mg<0u^{2}<5ga\ \Rightarrow\ T_{\text{top}}=\frac{mu^{2}}{a}-5mg<0

    A string would have gone slack: the required force is outwards, which a string cannot supply.

  3. Write the force in the rod as a function of the angle

    F=mu2a−2mg+3mgcos⁡θF=\frac{mu^{2}}{a}-2mg+3mg\cos\theta

    The energy equation and the radial equation together eliminate vv.

  4. Find where the force changes sign

    F=0 ⇒ cos⁡θ=2ga−u23gaF=0\ \Rightarrow\ \cos\theta=\frac{2ga-u^{2}}{3ga}

    With 4ga<u2<5ga4ga<u^{2}<5ga this lies strictly between −1-1 and −23-\frac{2}{3}, so the change happens near the top.

  5. Interpret the sign of FF beyond that angle

    F<0 (a thrust)F<0\ \text{(a thrust)}

    Beyond that point the rod must push PP outwards to keep it on the circle.

  6. State the acceleration of a particle moving on a circle

    aradial=v2ra_{\text{radial}}=\frac{v^{2}}{r}

    A particle moving on a circle of radius rr with speed vv has an acceleration of magnitude v2r\frac{v^{2}}{r} directed towards the centre.

  7. Write down Newton's second law towards the centre

    Ftowards O=mv2aF_{\text{towards }O}=\frac{mv^{2}}{a}

    Only the component of the resultant force along POPO produces the circular motion.

  8. Note that the tension does no work

    T⋅v=0\mathbf{T}\cdot\mathbf{v}=0

    The tension acts along POPO while the velocity is along the tangent, so the mechanical energy of PP is conserved.

  9. State conservation of energy on the circle

    12mu2=12mv2+mgh\frac{1}{2}mu^{2}=\frac{1}{2}mv^{2}+mgh

    The weight is the only force that does work, so the total mechanical energy is constant.

  10. State the height risen above the lowest point

    h=a(1−cos⁡θ)h=a\left(1-\cos\theta\right)

    With θ\theta measured from the downward vertical, PP lies a distance acos⁡θa\cos\theta below OO.

  11. Note that the mass cancels in the energy equation

    12u2=12v2+ga(1−cos⁡θ)\frac{1}{2}u^{2}=\frac{1}{2}v^{2}+ga\left(1-\cos\theta\right)

    Every term carries a factor mm, so the speed at a given point does not depend on the mass.

  12. Recall that a string can only pull

    T≥0T\geq 0

    A string may go slack, but it can never push the particle outwards.

  13. Recall the condition for complete circles on a string

    u2≥5gau^{2}\geq 5ga

    The tension must stay non-negative all the way to the highest point.

  14. Recall the condition at the highest point for a string

    vtop2≥gav_{\text{top}}^{2}\geq ga

    At the top the weight alone must not exceed the force needed to hold PP on the circle.

  15. Recall that a rod can push as well as pull

    T<0 is allowed (a thrust)T<0\ \text{is allowed (a thrust)}

    A rigid rod can exert a thrust, so the only requirement is that PP reaches the top.

  16. Select the correct description

    cos⁡θ=2ga−u23ga\cos\theta=\frac{2ga-u^{2}}{3ga}

    The rod carries a thrust from this angle up to the highest point.

Answer
PP performs complete vertical circles, but near the top the rod exerts a thrust rather than a tension; the force in the rod is zero where cos⁡θ=2ga−u23ga\cos\theta=\frac{2ga-u^{2}}{3ga}.
Question 2
9 markschallenging
A particle PP is projected horizontally with speed uu from the highest point of a fixed smooth solid sphere of radius aa and centre OO, and moves on the outer surface. Let θ\theta be the angle that OPOP makes with the upward vertical. Which of the following gives cos⁡θ\cos\theta at the instant when PP leaves the surface?
Show worked solution

Worked solution

  1. Apply conservation of energy from the highest point

    v2=u2+2ga(1−cos⁡θ)v^{2}=u^{2}+2ga\left(1-\cos\theta\right)

    The particle has fallen a height a(1−cos⁡θ)a\left(1-\cos\theta\right) below the top.

  2. Apply Newton's second law towards OO

    mgcos⁡θ−R=mv2amg\cos\theta-R=\frac{mv^{2}}{a}

    The normal reaction points away from the centre of the sphere.

  3. Set R=0R=0

    gacos⁡θ=v2ga\cos\theta=v^{2}

    Contact is lost the instant the normal reaction reaches zero.

  4. Substitute for v2v^{2} and solve

    gacos⁡θ=u2+2ga−2gacos⁡θ ⇒ cos⁡θ=u2+2ga3gaga\cos\theta=u^{2}+2ga-2ga\cos\theta\ \Rightarrow\ \cos\theta=\frac{u^{2}+2ga}{3ga}

    Setting u=0u=0 recovers the familiar cos⁡θ=23\cos\theta=\frac{2}{3}.

  5. State the acceleration of a particle moving on a circle

    aradial=v2ra_{\text{radial}}=\frac{v^{2}}{r}

    A particle moving on a circle of radius rr with speed vv has an acceleration of magnitude v2r\frac{v^{2}}{r} directed towards the centre.

  6. Write down Newton's second law towards the centre

    Ftowards O=mv2aF_{\text{towards }O}=\frac{mv^{2}}{a}

    Only the component of the resultant force along POPO produces the circular motion.

  7. Note that the tension does no work

    T⋅v=0\mathbf{T}\cdot\mathbf{v}=0

    The tension acts along POPO while the velocity is along the tangent, so the mechanical energy of PP is conserved.

  8. State conservation of energy on the circle

    12mu2=12mv2+mgh\frac{1}{2}mu^{2}=\frac{1}{2}mv^{2}+mgh

    The weight is the only force that does work, so the total mechanical energy is constant.

  9. State the height risen above the lowest point

    h=a(1−cos⁡θ)h=a\left(1-\cos\theta\right)

    With θ\theta measured from the downward vertical, PP lies a distance acos⁡θa\cos\theta below OO.

  10. Note that the mass cancels in the energy equation

    12u2=12v2+ga(1−cos⁡θ)\frac{1}{2}u^{2}=\frac{1}{2}v^{2}+ga\left(1-\cos\theta\right)

    Every term carries a factor mm, so the speed at a given point does not depend on the mass.

  11. Recall that a string can only pull

    T≥0T\geq 0

    A string may go slack, but it can never push the particle outwards.

  12. Recall the condition for complete circles on a string

    u2≥5gau^{2}\geq 5ga

    The tension must stay non-negative all the way to the highest point.

  13. Recall the condition at the highest point for a string

    vtop2≥gav_{\text{top}}^{2}\geq ga

    At the top the weight alone must not exceed the force needed to hold PP on the circle.

  14. Recall that a rod can push as well as pull

    T<0 is allowed (a thrust)T<0\ \text{is allowed (a thrust)}

    A rigid rod can exert a thrust, so the only requirement is that PP reaches the top.

  15. Select the expression

    cos⁡θ=u2+2ga3ga\cos\theta=\frac{u^{2}+2ga}{3ga}

    If u2≥gau^{2}\geq ga this is at least 11, meaning PP leaves the surface immediately.

Answer
cos⁡θ=u2+2ga3ga\cos\theta=\frac{u^{2}+2ga}{3ga}
Question 3
9 markschallenging
A particle PP of mass mm is attached to one end of a light rigid rod of length aa, freely pivoted at a fixed point OO, and is projected horizontally with speed uu from the point vertically below OO. In deriving the condition for complete vertical circles the force in the rod at the highest point has been written as F=mu2a−5mgF=\frac{mu^{2}}{a}-5mg. Which of the following correctly completes the argument?
Show worked solution

Worked solution

  1. Recall what a rod can do that a string cannot

    F<0 is allowedF<0\ \text{is allowed}

    A rigid rod can push the particle outwards, i.e. exert a thrust.

  2. Deduce that F≥0F\geq 0 is NOT the right condition

    F≥0 is the condition for a STRINGF\geq 0\ \text{is the condition for a STRING}

    Imposing it here would give the wrong answer, u2≥5gau^{2}\geq 5ga.

  3. State the correct condition

    vtop2≥0v_{\text{top}}^{2}\geq 0

    The particle simply has to arrive at the highest point.

  4. Use energy to express it in terms of uu

    vtop2=u2−4ga≥0 ⇒ u2≥4gav_{\text{top}}^{2}=u^{2}-4ga\geq 0\ \Rightarrow\ u^{2}\geq 4ga

    The top of the circle is a height 2a2a above the bottom.

  5. Note the strict inequality for a genuine complete circle

    u2>4gau^{2}>4ga

    With equality the particle arrives at the top with zero speed and stops there, so a strict inequality is needed to keep it going.

  6. State the acceleration of a particle moving on a circle

    aradial=v2ra_{\text{radial}}=\frac{v^{2}}{r}

    A particle moving on a circle of radius rr with speed vv has an acceleration of magnitude v2r\frac{v^{2}}{r} directed towards the centre.

  7. Write down Newton's second law towards the centre

    Ftowards O=mv2aF_{\text{towards }O}=\frac{mv^{2}}{a}

    Only the component of the resultant force along POPO produces the circular motion.

  8. Note that the tension does no work

    T⋅v=0\mathbf{T}\cdot\mathbf{v}=0

    The tension acts along POPO while the velocity is along the tangent, so the mechanical energy of PP is conserved.

  9. State conservation of energy on the circle

    12mu2=12mv2+mgh\frac{1}{2}mu^{2}=\frac{1}{2}mv^{2}+mgh

    The weight is the only force that does work, so the total mechanical energy is constant.

  10. State the height risen above the lowest point

    h=a(1−cos⁡θ)h=a\left(1-\cos\theta\right)

    With θ\theta measured from the downward vertical, PP lies a distance acos⁡θa\cos\theta below OO.

  11. Note that the mass cancels in the energy equation

    12u2=12v2+ga(1−cos⁡θ)\frac{1}{2}u^{2}=\frac{1}{2}v^{2}+ga\left(1-\cos\theta\right)

    Every term carries a factor mm, so the speed at a given point does not depend on the mass.

  12. Recall that a string can only pull

    T≥0T\geq 0

    A string may go slack, but it can never push the particle outwards.

  13. Recall the condition for complete circles on a string

    u2≥5gau^{2}\geq 5ga

    The tension must stay non-negative all the way to the highest point.

  14. Recall the condition at the highest point for a string

    vtop2≥gav_{\text{top}}^{2}\geq ga

    At the top the weight alone must not exceed the force needed to hold PP on the circle.

  15. Recall that a rod can push as well as pull

    T<0 is allowed (a thrust)T<0\ \text{is allowed (a thrust)}

    A rigid rod can exert a thrust, so the only requirement is that PP reaches the top.

  16. State the acceleration of a particle moving on a circle

    aradial=v2ra_{\text{radial}}=\frac{v^{2}}{r}

    A particle moving on a circle of radius rr with speed vv has an acceleration of magnitude v2r\frac{v^{2}}{r} directed towards the centre.

  17. Select the correct completion

    u2>4gau^{2}>4ga

    This is the condition for complete vertical circles on a rod.

Answer
FF may be negative, because a rod can exert a thrust, so the only requirement is that PP reaches the top: vtop2=u2−4ga≥0v_{\text{top}}^{2}=u^{2}-4ga\geq 0, giving u2>4gau^{2}>4ga.
Question 4
9 markschallenging
A smooth solid sphere of radius 1.6 m1.6\ \text{m} is fixed with its centre at a fixed point OO. A particle PP of mass 0.5 kg0.5\ \text{kg} is projected horizontally with speed u m s−1u\ \text{m}\ \text{s}^{-1} from the highest point of the sphere. Take g=9.8 m s−2g=9.8\ \text{m}\ \text{s}^{-2}. Find the least value of uu for which PP leaves the surface of the sphere immediately. Give your answer to 33 significant figures.
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Worked solution

  1. Apply Newton's second law at the highest point of the sphere

    mg−R=mu2amg-R=\frac{mu^{2}}{a}

    At the top the weight acts towards OO and the normal reaction away from OO; the acceleration towards OO is u2a\frac{u^{2}}{a}.

  2. Rearrange for the normal reaction

    R=m(g−u2a)R=m\left(g-\frac{u^{2}}{a}\right)

    The faster PP is projected, the smaller the reaction has to be.

  3. Impose R≤0R\leq 0 for PP to leave the surface at once

    g−u2a≤0 ⇒ u2≥gag-\frac{u^{2}}{a}\leq 0\ \Rightarrow\ u^{2}\geq ga

    The surface cannot pull, so once RR would have to be negative the particle has already left.

  4. Substitute the numbers

    u2≥9.8×1.6=15.68u^{2}\geq 9.8\times 1.6=15.68

    This is the least value of u2u^{2} for immediate departure.

  5. Interpret the condition physically

    u2a=g\frac{u^{2}}{a}=g

    The required centripetal acceleration at the top equals gg, so gravity is doing all of the work of bending the path.

  6. Test a speed just below the threshold

    u=3.92 ⇒ R>0u=3.92\ \Rightarrow\ R>0

    Just below the threshold the sphere is still pushing, so PP stays in contact for a while: the threshold is exactly right.

  7. Note what happens for a slower projection

    R>0 until cos⁡θ=u2+2ga3gaR>0\ \text{until}\ \cos\theta=\frac{u^{2}+2ga}{3ga}

    A slower particle stays on the surface until this angle is reached.

  8. State the acceleration of a particle moving on a circle

    aradial=v2ra_{\text{radial}}=\frac{v^{2}}{r}

    A particle moving on a circle of radius rr with speed vv has an acceleration of magnitude v2r\frac{v^{2}}{r} directed towards the centre.

  9. Write down Newton's second law towards the centre

    Ftowards O=mv2aF_{\text{towards }O}=\frac{mv^{2}}{a}

    Only the component of the resultant force along POPO produces the circular motion.

  10. Note that the tension does no work

    T⋅v=0\mathbf{T}\cdot\mathbf{v}=0

    The tension acts along POPO while the velocity is along the tangent, so the mechanical energy of PP is conserved.

  11. State conservation of energy on the circle

    12mu2=12mv2+mgh\frac{1}{2}mu^{2}=\frac{1}{2}mv^{2}+mgh

    The weight is the only force that does work, so the total mechanical energy is constant.

  12. State the height risen above the lowest point

    h=a(1−cos⁡θ)h=a\left(1-\cos\theta\right)

    With θ\theta measured from the downward vertical, PP lies a distance acos⁡θa\cos\theta below OO.

  13. Note that the mass cancels in the energy equation

    12u2=12v2+ga(1−cos⁡θ)\frac{1}{2}u^{2}=\frac{1}{2}v^{2}+ga\left(1-\cos\theta\right)

    Every term carries a factor mm, so the speed at a given point does not depend on the mass.

  14. Recall that a string can only pull

    T≥0T\geq 0

    A string may go slack, but it can never push the particle outwards.

  15. Recall the condition for complete circles on a string

    u2≥5gau^{2}\geq 5ga

    The tension must stay non-negative all the way to the highest point.

  16. State the least speed of projection

    umin⁡=15.68=3.96 m s−1u_{\min}=\sqrt{15.68}=3.96\ \text{m}\ \text{s}^{-1}

    At exactly this speed the weight alone provides the force needed to keep PP on the circle, and R=0R=0.

Answer
3.96 m s−13.96\ \text{m}\ \text{s}^{-1}
Question 5
9 markschallenging
A particle PP of mass 0.5 kg0.5\ \text{kg} moves on the inside of a smooth circular track of radius 0.7 m0.7\ \text{m} which is fixed in a vertical plane with centre OO. The particle passes through the lowest point of the track with speed u m s−1u\ \text{m}\ \text{s}^{-1}. The magnitude of the normal reaction of the track on PP at the lowest point of the track is four times the magnitude of the normal reaction at the highest point of the track. Take g=9.8 m s−2g=9.8\ \text{m}\ \text{s}^{-2}. Find the value of uu. Give your answer to 33 significant figures.
Show worked solution

Worked solution

  1. Write the force at the lowest point

    R1=mu2a+mgR_{1}=\frac{mu^{2}}{a}+mg

    At the lowest point the force acts towards OO and the weight away from OO.

  2. Write the force at the highest point

    R2=mvtop2a−mgR_{2}=\frac{mv_{\text{top}}^{2}}{a}-mg

    At the top both the force and the weight act towards OO, so R2+mg=mvtop2aR_{2}+mg=\frac{mv_{\text{top}}^{2}}{a}.

  3. Use energy to write vtop2v_{\text{top}}^{2} in terms of uu

    vtop2=u2−4gav_{\text{top}}^{2}=u^{2}-4ga

    The highest point is a height 2a2a above the lowest point.

  4. Express the force at the top in terms of uu

    R2=mu2a−5mgR_{2}=\frac{mu^{2}}{a}-5mg

    Substituting the energy result removes vtopv_{\text{top}}.

  5. Form the given relation

    mu2a+mg=4(mu2a−5mg)\frac{mu^{2}}{a}+mg=4\left(\frac{mu^{2}}{a}-5mg\right)

    The mass cancels from every term.

  6. Solve for u2u^{2}

    u2=21ga3 ⇒ u2=48.02u^{2}=\frac{21ga}{3}\ \Rightarrow\ u^{2}=48.02

    Collecting the terms in u2u^{2} gives a linear equation.

  7. Check the forces with this value of uu

    R1=39.2 N,R2=9.8 NR_{1}=39.2\ \text{N},\qquad R_{2}=9.8\ \text{N}

    The first is exactly 44 times the second, as required.

  8. Check the standard difference of the extreme forces

    R1−R2=6mg=29.4 NR_{1}-R_{2}=6mg=29.4\ \text{N}

    The difference between the greatest and least values is always 6mg6mg.

  9. Check that complete circles are possible

    u2=48.02≥5ga=34.3u^{2}=48.02\geq 5ga=34.3

    The particle does complete the circle, so the force at the top is meaningful.

  10. State the acceleration of a particle moving on a circle

    aradial=v2ra_{\text{radial}}=\frac{v^{2}}{r}

    A particle moving on a circle of radius rr with speed vv has an acceleration of magnitude v2r\frac{v^{2}}{r} directed towards the centre.

  11. Write down Newton's second law towards the centre

    Ftowards O=mv2aF_{\text{towards }O}=\frac{mv^{2}}{a}

    Only the component of the resultant force along POPO produces the circular motion.

  12. Note that the tension does no work

    T⋅v=0\mathbf{T}\cdot\mathbf{v}=0

    The tension acts along POPO while the velocity is along the tangent, so the mechanical energy of PP is conserved.

  13. State conservation of energy on the circle

    12mu2=12mv2+mgh\frac{1}{2}mu^{2}=\frac{1}{2}mv^{2}+mgh

    The weight is the only force that does work, so the total mechanical energy is constant.

  14. State the height risen above the lowest point

    h=a(1−cos⁡θ)h=a\left(1-\cos\theta\right)

    With θ\theta measured from the downward vertical, PP lies a distance acos⁡θa\cos\theta below OO.

  15. State the speed of projection

    u=48.02=6.93 m s−1u=\sqrt{48.02}=6.93\ \text{m}\ \text{s}^{-1}

    This is the speed of PP at the lowest point.

Answer
6.93 m s−16.93\ \text{m}\ \text{s}^{-1}

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